LeetCode Unique Binary Search Trees 2 递归方法

//递归方法解决
//I start by noting that 1..n is the in-order traversal for any BST with nodes 1 to n.
//So if I pick i-th node as my root, the left subtree will contain elements 1 to (i-1),
//and the right subtree will contain elements (i+1) to n.
//I use recursive calls to get back all possible trees for left and right subtrees
//and combine them in all possible ways with the root.

/************************************************************************************ Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.** For example,* Given n = 3, your program should return all 5 unique BST's shown below.**    1         3     3      2      1*     \       /     /      / \      \*      3     2     1      1   3      2**********************************************************************************/#include <stdio.h>#include <stdlib.h>#include <string.h>#include <vector>using namespace std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};vector<TreeNode*> generateTrees(int low, int high);vector<TreeNode*> generateTrees(int n) {    vector<TreeNode*> v;    if (n==0) return v;  //i find add this can ac ,if do not add it is return wrong.    v = generateTrees(1, n);      return v;}vector<TreeNode*> generateTrees(int low, int high) {    vector<TreeNode*> v;    if (low > high || low <= 0 || high <= 0) {        v.push_back(NULL);        return v;    }    if (low == high) {        TreeNode* node = new TreeNode(low);        v.push_back(node);        return v;    }    //the i is root,and construct the left subtree and right subtree.    for (int i = low; i <= high; i++) {        vector<TreeNode*> vleft = generateTrees(low, i - 1);        vector<TreeNode*> vright = generateTrees(i + 1, high);        for (int l = 0; l<vleft.size(); l++) {            for (int r = 0; r<vright.size(); r++) {                TreeNode *root = new TreeNode(i);                root->left = vleft[l];                root->right = vright[r];                v.push_back(root);            }        }    }    return v;}void printTree(TreeNode *root) {    if (root == NULL) {        printf("# ");        return;    }    printf("%d ", root->val);    printTree(root->left);    printTree(root->right);}int main(int argc, char** argv){    int n = 2;    vector<TreeNode*> v = generateTrees(n);    for (int i = 0; i<v.size(); i++) {        printTree(v[i]);        printf("\n");    }    return 0;}